What is the first quartile, Q1, of the data set?

Mathematics

1 answer:

Kryger [21]4 years ago

4 0

It is 28 because you have to take the median out then go to the first set of data and find the median of that.

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A line segment has endpoints (-2,-3) and (4,4). Find the length of the segment and find the midpoint.

uysha [10]

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ -2 &,& -3~) 
% (c,d)
&&(~ 4 &,& 4~)
\end{array}
\\\\\\
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
d=\sqrt{[4-(-2)]^2+[4-(-3)]^2}\implies d=\sqrt{(4+2)^2+(4+3)^2}
\\\\\\
d=\sqrt{36+49}\implies \boxed{d=\sqrt{85}}\\\\
-------------------------------$

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points }
\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ -2 &,& -3~) 
% (c,d)
&&(~ 4 &,& 4~)
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{ x_2 + x_1}{2}\quad ,\quad \cfrac{ y_2 + y_1}{2} \right)
\\\\\\
\left( \cfrac{4-2}{2}~~,~~\cfrac{4-3}{2} \right)\implies \left(\cfrac{2}{2}~~,~~\cfrac{1}{2} \right)\implies \boxed{\left(1~,~\frac{1}{2} \right)}$

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4 years ago

Maxine spent 15 hours doing her homework last week. This week she spent 18 hours doing homework. She says that she spent 120% mo

KengaRu [80]

The answer is true because 18 is bigger then 15

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4 years ago

Sin 2x + tan 2x = 4tan/ 1- tan^4 prove that

nalin [4]

$\dfrac{4\tan x}{1-\tan^4x}=\dfrac{4\tan x}{(1-\tan^2x)(1+\tan^2x)}$
$=\dfrac{4\tan x}{(1-(\sec^2x-1)\sec^2x}$
$=\dfrac{4\tan x\cos^2x}{2-\sec^2x}$
$=\dfrac{4\sin x\cos x}{2-\sec^2x}$
$=\dfrac{4\sin x\cos^3x}{2\cos^2x-1}$
$=\dfrac{2\sin2x\cos^2x}{\cos2x}$
$=2\tan2x\cos^2x$
$=2\tan2x\left(\dfrac{1+\cos2x}2\right)$
$=\tan2x(1+\cos2x)$
$=\tan2x+\sin2x$

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3 years ago

What does 270 and 3000 have in common

FromTheMoon [43]

They are both multiples of 3

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4 years ago

Read 2 more answers

Calculate y as a function of x when dy/dx = 4x3 + 3x2 - 6x + 5

PtichkaEL [24]

Answer:
y = x⁴ + x³ - 3x² + 5x + C

======

Separable differential equations such as these ones can be solved by treating dy/dx as a ratio of differentials. Then move the dx with all the x terms and move the dy with all the y terms. After that, integrate both sides of the equation.

$\begin{aligned}
\dfrac{dy}{dx} &= 4x^3 + 3x^2 - 6x + 5 \\
dy &= (4x^3 + 3x^2 - 6x + 5) dx \\
\int dy &= \int (4x^3 + 3x^2 - 6x + 5) dx 
\end{aligned}$

In general (understood that +C portions are still there),

$\int x^{m} = \dfrac{x^{m+1}}{m+1}$

Note that ∫dy = y since it is ∫1·dy = ∫y⁰ dy = y¹/(0+1) = y
For the right-hand side, we use the sum/difference rule for integrals, which says that

$\int \big[f(x) \pm g(x)\big]\, dx = \int f(x)\,dx \pm \int g(x) \, dx$

Applying these concepts:

$\begin{aligned} 
 \int dy &= \int (4x^3 + 3x^2 - 6x + 5) \, dx \\
y &= \int 4x^3\,dx + \int 3x^2 \, dx - \int 6x\, dx + \int 5\, dx \\
&= \frac{4x^4}{4} + \frac{3x^3}{3} - \frac{6x^2}{2} + 5x + C \qquad \text{(only one $C$ is needed)} \\
&= x^4 + x^3 - 3x^2 + 5x + C
\end{aligned}$

The answer is y = x⁴ + x³ - 3x² + 5x + C

6 0

3 years ago